Statement - 1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16)+...+(361 + 380 + 400) is 8000.
Statement - 2: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3},$ for any natural number n.

Statement - 1 is false, Statement - 2 is true.

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Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1.

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Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1.

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Statement - 1 is true, Statement - 2 is false.

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Solution

The correct option is B Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1.
$n^{t h}$ term of the given series
$T_{n} = (n - 1)^{2} + (n - 1) n + n^{2}$
$= \frac{((n - 1)^{3} - n^{3})}{(n - 1) - n} = n^{3} - (n - 1)^{3} \Rightarrow S_{n} = \sum_{k = 1}^{n} [k^{3} - (k - 1)^{3}] \Rightarrow 8000 = n^{3}$
$\Rightarrow n = 20$ which is a natural number
$T_{1} = 1^{3} - 0^{3} T_{2} = 2^{3} - 1^{3} ⋮ T_{20} = 20^{3} - 19^{3}$
Now, $T_{1} + T_{2} + . . . + T_{20} = S_{20} \Rightarrow S_{20} = 20^{3} - 0^{3} = 8000$
Hence, both the given statements are true and statement 2 supports statement 1.

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\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3},

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\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3},

1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)

8000.

n \sum k = 1 (k^{3} - (k - 1)^{3}) = n^{3}

n

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